Answer
$\sqrt {33}; \lt \dfrac{8}{\sqrt {33}},\dfrac{-2}{\sqrt {33}},\dfrac{8}{\sqrt {33}}\gt$
Work Step by Step
As we know that the unit vector $\hat{\textbf{u}}$ is defined as: $\hat{\textbf{u}}=\dfrac{v}{|v|}$
Now, we have: $|v|=\sqrt{(4)^2+(-1)^2+(4)^2}=\sqrt {33}$
Thus, $2 \hat{\textbf{u}}=\dfrac{v}{|v|}=2 [\dfrac{\lt 4,-1,4 \gt}{\sqrt {33}}]= \lt \dfrac{8}{\sqrt {33}},\dfrac{-2}{\sqrt {33}},\dfrac{8}{\sqrt {33}}\gt$
